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A group of four women and three men have tickets for seven [#permalink]
15 Aug 2007, 18:17

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A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!(B) 7! – 4!3!(C) 7! – 5!3!(D) 7 × 2!3!2!(E) 2!3!2!

going to try to work it out even though I am probably wrong different possibilities - we know that the men and women will alternate seats since men will not sit next to each other
1) M W M W M W W

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!(B) 7! – 4!3!(C) 7! – 5!3!(D) 7 × 2!3!2!(E) 2!3!2!

C. = 7! - 6x5x4!= 7! - 6!= 7! - 5!3x2= 7! - 5!3!

can you explain this bit.

my way is
total ways =7!
ways that men can sit together with women=5! ways. considering men as a single seat.
3!= no. of ways men can sit together.

A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! – 2!3!2!(B) 7! – 4!3!(C) 7! – 5!3!(D) 7 × 2!3!2!(E) 2!3!2!

C. = 7! - 6x5x4!= 7! - 6!= 7! - 5!3x2= 7! - 5!3!

can you explain this bit.

my way is total ways =7!ways that men can sit together with women=5! ways. considering men as a single seat. 3!= no. of ways men can sit together.

So 7!- 5!.3!.

required = total - 3 men adjecent

total = 7!
3 men adjacent to each other can be placed in 5 places = 5
3 men can be placed adjacent to each other in a place = 3! = 3x2 = 6 ways
remaining 4 people can be placed = 4! ways